Let me define the $L^p$ space.
Let $X$ be a measure space and $1\leq p \leq \infty$ then the usual $L^p$ space is
$$L^p(X)=\left\{f : X\to \mathbb{R} \text{ such that } \int_{X}|f|^{p}d\mu < \infty \right\}$$
We define the convex set by letting $S$ be a vector space over real numbers (or some ordered field); then a subset $C$ of $S$ is said to be a convex set if $\forall x,y\in C$ we have $\alpha x+\beta y \in C$ where $\alpha + \beta =1$
So suppose if I choose two functions $f,g \in L^p$ and some $\alpha , \beta$ such that $\alpha + \beta =1$; then we can see that $\alpha f + \beta g \in L^p$ since $$\int_{X}(|\alpha f + \beta g| \leq \int_{X}|\alpha f| + \int_{X} |\beta g| \leq \alpha\int_{X}|f|+\beta \int_{X}|g|$$
Am I right or I am making some mistake? Please correct me if there is some mistake; according to me $L^p$ is a convex set but I am not sure.
You're right (in the case of $L^1$), but this ultimately has nothing to do with $L^p$ spaces.
Take $V$ an $F$-vector space. (For simplicity, we'll let $F \in \{\mathbb{R,C}\}$.) Then for any $v \in V$ and $a \in F$, $av$ exists and lies in $F$. Same with addition: for $v,w \in V$, then $v+W \in F$.
Consequently, naturally, for $a,b \in F$ and $v,w \in V$, $av + bw \in V$ -- this includes the narrower case of $a+b=1$.
Of course, then, any vector space over $\mathbb{R}$ or $\mathbb{C}$ is a convex subset of itself.
(That's the key thing to take away here: convexity is only really a nontrivial thing to be concerned about when it's a proper subset of the original space.)